Practice questions

(a) From the left, the graph approaches the open circle at height ${3}$:

$\displaystyle\lim_{x\to 2^-}f(x)=3$

(b) From the right, the graph also approaches the open circle at height ${3}$:

$\displaystyle\lim_{x\to 2^+}f(x)=3$

(c) The two one-sided limits are equal. Both are ${3}$. So

$\displaystyle\lim_{x\to 2}f(x)=3$

(d) The solid dot at ${x=2}$ is at height ${1}$:

${f(2)=1}$

The limit is ${3}$, but the value is ${1}$. The value at the point does not change the limit.

(e) From the left, the graph approaches the open circle at height ${1}$:

$\displaystyle\lim_{x\to 4^-}f(x)=1$

(f) From the right, the graph starts at the solid dot at height ${3}$:

$\displaystyle\lim_{x\to 4^+}f(x)=3$

(g) The two one-sided limits are different (${1}$ and ${3}$). So $\displaystyle\lim_{x\to 4}f(x)$ does not exist.

(h) The solid dot at ${x=4}$ is at height ${3}$:

${f(4)=3}$

For ${x<3}$, use ${x^2-1}$:

$\displaystyle\lim_{x\to 3^-}f(x)=3^2-1=8$

For ${x>3}$, use ${2x+2}$:

$\displaystyle\lim_{x\to 3^+}f(x)=2(3)+2=8$

The two one-sided limits are equal. So

$\displaystyle\lim_{x\to 3}f(x)=8$

For ${x>2}$, the number ${x-2}$ is positive. So ${|x-2|=x-2}$, and

${\dfrac{|x-2|}{x-2}=\dfrac{x-2}{x-2}=1}$

So the right-hand limit is

$\displaystyle\lim_{x\to 2^+}\dfrac{|x-2|}{x-2}=1$

For ${x<2}$, the number ${x-2}$ is negative. So ${|x-2|=-(x-2)}$, and

$\dfrac{|x-2|}{x-2}=\dfrac{-(x-2)}{x-2}=-1$

So the left-hand limit is

$\displaystyle\lim_{x\to 2^-}\dfrac{|x-2|}{x-2}=-1$

The two one-sided limits are different. So the limit does not exist.

This is a polynomial. So use direct substitution: put in ${x=1}$.

$\begin{align*}&\lim_{x\to 1}(2x^2-x+3)\\&=2(1)^2-1+3\\&=4\end{align*}$

Direct substitution gives ${0/0}$. Factor the numerator:

$\begin{align*}&\lim_{x\to 2}\dfrac{x^2-5x+6}{x-2}\\&=\lim_{x\to 2}\dfrac{(x-2)(x-3)}{x-2}\end{align*}$

Cancel the common factor ${x-2}$, and substitute ${x=2}$:

$\begin{align*}&=\lim_{x\to 2}(x-3)\\&=2-3\\&=-1\end{align*}$

Direct substitution gives ${0/0}$. Factor the numerator and the denominator:

$\begin{align*}&\lim_{x\to 1}\dfrac{x^2+2x-3}{x^2-1}\\&=\lim_{x\to 1}\dfrac{(x+3)(x-1)}{(x+1)(x-1)}\end{align*}$

Cancel the common factor ${x-1}$, and substitute ${x=1}$:

$\begin{align*}&=\lim_{x\to 1}\dfrac{x+3}{x+1}\\&=\dfrac{1+3}{1+1}\\&=2\end{align*}$

Direct substitution gives ${0/0}$. Multiply the numerator and the denominator by the conjugate, ${\sqrt{x+4}+2}$:

$\begin{align*}&\lim_{x\to 0}\dfrac{\sqrt{x+4}-2}{x}\\&=\lim_{x\to 0}\dfrac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)}\\&=\lim_{x\to 0}\dfrac{(x+4)-4}{x(\sqrt{x+4}+2)}\\&=\lim_{x\to 0}\dfrac{x}{x(\sqrt{x+4}+2)}\end{align*}$

Cancel ${x}$, and substitute ${x=0}$:

$\begin{align*}&=\lim_{x\to 0}\dfrac{1}{\sqrt{x+4}+2}\\&=\dfrac{1}{\sqrt{4}+2}\\&=\dfrac{1}{4}\end{align*}$

Direct substitution gives ${0/0}$. First, combine the two fractions in the numerator. Their common denominator is ${3(x+3)}$:

$\begin{align*}&\dfrac{1}{x+3}-\dfrac{1}{3}\\&=\dfrac{3-(x+3)}{3(x+3)}\\&=\dfrac{-x}{3(x+3)}\end{align*}$

Now divide by ${x}$:

$\begin{align*}&\lim_{x\to 0}\dfrac{\dfrac{1}{x+3}-\dfrac{1}{3}}{x}\\&=\lim_{x\to 0}\dfrac{-x}{3x(x+3)}\end{align*}$

Cancel ${x}$, and substitute ${x=0}$:

$\begin{align*}&=\lim_{x\to 0}\dfrac{-1}{3(x+3)}\\&=\dfrac{-1}{3(0+3)}\\&=-\dfrac{1}{9}\end{align*}$

Direct substitution gives ${0/0}$. Expand ${(2+h)^2}$:

$\begin{align*}&\lim_{h\to 0}\dfrac{(2+h)^2-4}{h}\\&=\lim_{h\to 0}\dfrac{4+4h+h^2-4}{h}\\&=\lim_{h\to 0}\dfrac{4h+h^2}{h}\\&=\lim_{h\to 0}\dfrac{h(4+h)}{h}\end{align*}$

Cancel ${h}$, and substitute ${h=0}$:

$\begin{align*}&=\lim_{h\to 0}(4+h)\\&=4\end{align*}$

The numerator is ${3}$, which is positive.

From the right (${x>1}$), the denominator ${x-1}$ is a small positive number. Positive over positive is positive. So

$\displaystyle\lim_{x\to 1^+}\dfrac{3}{x-1}=\infty$

From the left (${x<1}$), the denominator ${x-1}$ is a small negative number. Positive over negative is negative. So

$\displaystyle\lim_{x\to 1^-}\dfrac{3}{x-1}=-\infty$

The highest power of ${x}$ in the denominator is ${x^2}$. Divide every term by ${x^2}$:

$\begin{align*}&\lim_{x\to\infty}\dfrac{4x^2-3x}{2x^2+1}\\&=\lim_{x\to\infty}\dfrac{4-\dfrac{3}{x}}{2+\dfrac{1}{x^2}}\\&=\dfrac{4-0}{2+0}\\&=2\end{align*}$

The highest power of ${x}$ in the denominator is ${x^2}$. Divide every term by ${x^2}$:

$\begin{align*}&\lim_{x\to\infty}\dfrac{5x+2}{x^2-3}\\&=\lim_{x\to\infty}\dfrac{\dfrac{5}{x}+\dfrac{2}{x^2}}{1-\dfrac{3}{x^2}}\\&=\dfrac{0+0}{1-0}\\&=0\end{align*}$

Vertical asymptote: The denominator ${x-3}$ is ${0}$ at ${x=3}$. The numerator there is ${2(3)+1=7}$, which is not ${0}$. So the function becomes infinite near ${x=3}$, and

${x=3}$

is a vertical asymptote.

Horizontal asymptote: Divide every term by ${x}$:

$\begin{align*}&\lim_{x\to\infty}\dfrac{2x+1}{x-3}\\&=\lim_{x\to\infty}\dfrac{2+\dfrac{1}{x}}{1-\dfrac{3}{x}}\\&=\dfrac{2+0}{1-0}\\&=2\end{align*}$

The limit as ${x\to -\infty}$ is also ${2}$, in the same way. So

${y=2}$

is a horizontal asymptote.

We cannot substitute ${x=0}$, because ${\dfrac{1}{x}}$ is not defined there. Use the Squeeze Theorem.

The cosine of any number is between ${-1}$ and ${1}$:

${-1\le\cos\left(\dfrac{1}{x}\right)\le 1}$

Multiply each part by ${x^2}$. Since ${x^2\ge 0}$, the inequality signs do not change:

$-x^2\le x^2\cos\left(\dfrac{1}{x}\right)\le x^2$

The two outside functions both approach ${0}$:

$\displaystyle\lim_{x\to 0}(-x^2)=0$   and   $\displaystyle\lim_{x\to 0}x^2=0$

The middle function is trapped between them. So, by the Squeeze Theorem,

$\displaystyle\lim_{x\to 0}x^2\cos\left(\dfrac{1}{x}\right)=0$

To use ${\sin u/u\to 1}$, the angle and the denominator must match. Here the angle is ${4x}$. So multiply the numerator and the denominator by ${4}$:

$\begin{align*}&\lim_{x\to 0}\dfrac{\sin 4x}{x}\\&=\lim_{x\to 0}4\cdot\dfrac{\sin 4x}{4x}\end{align*}$

Let ${u=4x}$. As ${x\to 0}$, also ${u\to 0}$. So

$\begin{align*}&=4\lim_{u\to 0}\dfrac{\sin u}{u}\\&=4\cdot 1\\&=4\end{align*}$

The angle is ${2x}$, but the denominator is ${3x}$. Write the denominator with ${2x}$:

$\begin{align*}&\lim_{x\to 0}\dfrac{\sin 2x}{3x}\\&=\lim_{x\to 0}\dfrac{2}{3}\cdot\dfrac{\sin 2x}{2x}\end{align*}$

Let ${u=2x}$. As ${x\to 0}$, also ${u\to 0}$. So

$\begin{align*}&=\dfrac{2}{3}\lim_{u\to 0}\dfrac{\sin u}{u}\\&=\dfrac{2}{3}\cdot 1\\&=\dfrac{2}{3}\end{align*}$

At ${x=2}$: From Question 1, the limit is ${3}$, but ${f(2)=1}$. The limit and the value are different. So ${f}$ is not continuous at ${x=2}$.

The limit exists, so this is a removable discontinuity. We could fix it by moving the dot to height ${3}$.

At ${x=4}$: The left-hand limit is ${1}$, and the right-hand limit is ${3}$. So the limit does not exist, and ${f}$ is not continuous at ${x=4}$.

The graph jumps from height ${1}$ to height ${3}$. This is a jump discontinuity.

For continuity at ${x=3}$, the two one-sided limits must be equal. (They will then also equal ${g(3)=9+k}$.)

$\displaystyle\lim_{x\to 3^-}g(x)=3k-1$

$\displaystyle\lim_{x\to 3^+}g(x)=3^2+k=9+k$

Set them equal and solve for ${k}$:

$\begin{align*}3k-1&=9+k\\2k&=10\\k&=5\end{align*}$

So ${g}$ is continuous at ${x=3}$ when ${k=5}$.

This is a rational function. It is continuous everywhere except where the denominator is ${0}$. Factor the denominator:

${x^2-x-6=(x-3)(x+2)}$

It is ${0}$ at ${x=3}$ and at ${x=-2}$. So ${h}$ is continuous at every ${x}$ except ${-2}$ and ${3}$. In interval notation:

${(-\infty,-2)\cup(-2,3)\cup(3,\infty)}$

To find the type of each discontinuity, cancel the common factor ${x-3}$. For ${x\ne 3}$,

$h(x)=\dfrac{x-3}{(x-3)(x+2)}=\dfrac{1}{x+2}$

At ${x=3}$: The limit exists:

$\displaystyle\lim_{x\to 3}h(x)=\dfrac{1}{3+2}=\dfrac{1}{5}$

But ${h(3)}$ is not defined. So this is a removable discontinuity.

At ${x=-2}$: The denominator ${x+2}$ approaches ${0}$, but the numerator is ${1}$. So ${h(x)}$ becomes infinite. This is an infinite discontinuity. The line ${x=-2}$ is a vertical asymptote.

Move everything to one side. Let

${f(x)=\cos x-x}$.

A solution of ${\cos x=x}$ is a number where ${f(x)=0}$.

The functions ${\cos x}$ and ${x}$ are continuous, so their difference ${f}$ is continuous on ${\left[0,\dfrac{\pi}{2}\right]}$.

${f(0)=\cos 0-0=1}$, which is positive.

$f\left(\dfrac{\pi}{2}\right)=\cos\dfrac{\pi}{2}-\dfrac{\pi}{2}=-\dfrac{\pi}{2}$, which is negative.

The number ${0}$ is between these two values. So, by the Intermediate Value Theorem, there is a number ${c}$ between ${0}$ and ${\dfrac{\pi}{2}}$ with ${f(c)=0}$. That is, ${\cos c=c}$.

(A calculator shows that ${c\approx 0.739}$.)